Problem
GEO-B2-M06-P020 Recovering Concurrence
#20
★★★★☆ Level 4 of 5
In triangle \(ABC\), points \(D\in BC\), \(E\in CA\). Line \(DE\) meets the extension of \(AB\) at \(X\). Point \(F\in AB\) is chosen so that \(\frac{AF}{FB}=\frac{AX}{XB}\). Prove that \(AD\), \(BE\), \(CF\) are concurrent.
First apply Menelaus to \(D,E,X\), then replace \(AX:XB\) by \(AF:FB\).
By Menelaus for line \(D E X\): \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AX}{XB}=1\). By the condition, \(\frac{AX}{XB}=\frac{AF}{FB}\), so \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). By Ceva's theorem, lines \(AD\), \(BE\), \(CF\) are concurrent.
A problem on an auxiliary point and the transition Menelaus -> Ceva.